If $4sin\theta +3cosec\theta =4\sqrt{3}$, where θ is an acute angle (between 0° and 90°), then find the value of angle θ.

The correct answer is 60°.

Step-by-step Explanation:

We are given the trigonometric equation:

$4sin\theta +3cosec\theta =4\sqrt{3}$

Recall the reciprocal trigonometric identity: $cosec\theta =\frac{1}{sin\theta }$.

Substituting this identity into the equation gives:

$4sin\theta +\frac{3}{sin\theta }=4\sqrt{3}$

Let $sin\theta =x$. The equation simplifies to:

$4x+\frac{3}{x}=4\sqrt{3}$

Multiplying the entire equation by $x$ to clear the fraction:

$4x^2+3=4\sqrt{3}x$

Rearranging into standard quadratic equation form $a{x}^2+bx+c=0$:

$4x^2-4\sqrt{3}x+3=0$

This expression is a perfect square polynomial of the form ${\left(2x-\sqrt{3}\right)}^2$:

${\left(2x-\sqrt{3}\right)}^2=0$

Taking the square root of both sides:

$2x-\sqrt{3}=0$

$2x=\sqrt{3}$

$x=\frac{\sqrt{3}}{2}$

Substituting back $x=sin\theta$:

$sin\theta =\frac{\sqrt{3}}{2}$

Since θ is an acute angle (0° < θ < 90°), we find:

$\theta =\mathrm{60°}$